m*a_2 = 4*(pi)^2*m*a

ma_2 = 4(pi)^2ma_1 = G-force

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brian a m stuckless - 22 Sep 2005 15:46 GMT

Bilge wrote:

> The the entire point of the experiment is discover whether
> or not the `m' in Force = ma is the same `m' in Gm/r^2. --

You CONFUSE because your "Gm/r^2" is NOT a FORCE, Dimwit.!!
But, CLEANing up here, let notation for F = m2*a_2 ;

THEN: G FORCE F = m1*g1 = m1*v1^2 / r1
= G*M1*m1 / r1^2
= 4*(pi)^2*m1*r1 / t1^2
= 4*(pi)^2*m1*a_1
= m2*a_2.
CONCLUSiON:
GiVEN the mass m = m1 ..as per the Equivalence Principle;
AND ALSO the GiVEN FORCEs F are BOTH equivalent, as well;
THEN the ACCELERATiON a_2 ..is CLEARLY = to 4*(pi)^2*a_1.

The ACCELERATiON a_2, is CLEARLY equal to 4*(pi)^2*a_1.!!

CONVERSELY if a_1 gets "PEGGED" as equal to a_2, HOWEVER,
THEN mass m1 will HAVE to be EQUAL to mass 4*(pi)^2*m2.!!

GUESS, iNHERENTLY, does NOT have this 4*(pi)^2 PROBLEM.!!
Sincerely c,
```Brian

>><> >><> >><> >><> >><>

> -- A non-null
> result (which is what uncleal would like to get) means those
> two `m's are not the same. It shouldn't be much of a surprise
> that one can take advantage of the difference.
>
> >I.e. is the binding
> >energy of the system dependent on gravitational or intertial mass?
>
> The binding energy depends on the electromagnetic forces.
> Besides, I _dont_ need to use both isomers. It's sufficient to
> just disassemble the molecule, transport the constituent atoms
> and resassemble it.
>
> >In this gedanken, the gravitational mass of step (1) is greater than
> >after step (3).
>
> Not necessarily. I made no assumption about which isomer had
> a greater mass.
> >If it costs energy to put the object into the step (3)
> >configuration there is no problem with energy conservation. What if
> >the inertial mass is greater than the gravitational mass, and the
> >binding energy is tied to inertial mass?
>
> What if the sky falls? Don't you think any of these things occured to
> me before I ever posted that? Seriously, rather than just saying
> ``what if,'' without having any idea whether or not I already considered
> these things, find a particular ``what if'' and make some numerical
> estimates within the constraints we already know from experimental data.

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brian a m stuckless - 22 Sep 2005 18:13 GMT

SORRY Bilge ..about last POST.!! Bilge wrote:..

> The the entire point of the experiment is discover whether
> or not the `m' in Force = ma is the same `m' in Gm/r^2. --

You CONFUSE because "Gm/r^2" is NOT a FORCE, Bilge.!!

FOOT-in-MOUTH is the MOST GRUEsome afFLiCTion.. .. .. .. ..
..(RED-FACED) CLEANing up: Let notation for F = m2*a_2 ;

THEN: G-FORCE F = m1*g1
= m1*v1^2 / r1
= G*M1*m1 / r1^2
= 4*(pi)^2*m1*r1 / t1^2
= 4*(pi)^2*m2*(ORBiT radius) / (PERiOD sec)^2
= m2*(DiSTANCE) / (TiME ...second)^2
= m2*a_2.
CONCLUSiON:
An *EMBARASSiNG* example of "PEGGiNG" a "mis-nomer".!!

HANGin' up m'HOLSTER ( ..on that SleeveeEEN BilGE ).!!

VERY Sincerely c,
```Brian
p.s.
Note ANGULAR momentum pA = mass*radius*velocity:
But THERE's NO QUANTiTY for a CENTRAL MASS here.!!

Again, i am still red-faced about that misnomer.!!
That SLOPPY confsed Bilge makes me so irritated.!!

>><> >><> >><> >><> >><>

> -- A non-null
> result (which is what uncleal would like to get) means those
> two `m's are not the same. It shouldn't be much of a surprise
> that one can take advantage of the difference.

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Natural Science Forum / Physics / Particle Physics / September 2005

m*a_2 = 4*(pi)^2*m*a_1 = G-force

ma_2 = 4(pi)^2ma_1 = G-force